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ykannan's avatar
ykannan
Icon for Helper I rankHelper I
3 years ago
Solved

How to show Average values in Total section

I am trying to create PowerBI report with same data as in other BI tool,

In PowerBI,I have created measures like,

DO_Count = count(creo_sales[DO_ID])
 
EGPS DO Count = COUNTROWS (
    FILTER (
        creo_sales,
         CONTAINSSTRING(creo_sales[OP_GROUP_NAME], "GPS")))
 
Percent EGPS DO's = [EGPS DO Count]/[DO_Count]
 
In the percentage EGPS DO's columns, total section, I want to have Average value instaed of Total. In those above measures, I could not see Average aggregation.

 

 

Other BI tool,

DO Count= =Count([Delivered Order ID])

 

EGPS DO Count=Count([Delivered Order ID];Distinct) Where(Match([Order Planning Group];"*Excelsius GPS*")) ForEach([Delivered Order ID]) In([YEAR])

 

Percent EGPS DO's = =[EGPS DO Count]/[DO Count]

 

I can do the Average in other BI tool. How can I do the same in Power BI?

 

 

As per below article, Measures do not have the aggregate values below shown options,

Work with aggregates (sum, average, and so on) in Power BI - Power BI | Microsoft Learn

Thanks,

Yoga

Need Help
  • Anonymous's avatar
    Anonymous
    3 years ago

    Hi ykannan ,

     

    I suggest you to use Averagex() function to create a new measure based on [Percent EGPS DO's] measure.

    New Percent EGPS DO's =
AVERAGEX ( VALUES ( 'TableName'[Year] ), [Percent EGPS DO's] )

     

    Best Regards,
    Rico Zhou

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

     

1 Reply

  • Anonymous's avatar
    Anonymous
    Not applicable
    3 years ago

    Hi ykannan ,

     

    I suggest you to use Averagex() function to create a new measure based on [Percent EGPS DO's] measure.

    New Percent EGPS DO's =
AVERAGEX ( VALUES ( 'TableName'[Year] ), [Percent EGPS DO's] )

     

    Best Regards,
    Rico Zhou

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.